Internal problem ID [9035]
Internal file name [OUTPUT/7970_Monday_June_06_2022_01_03_06_AM_59801867/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 701.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime }-\frac {2 x \,{\mathrm e}^{x}-2 x -\ln \left (x \right )-1+\ln \left (x \right ) x^{4}+x^{4}-2 y x^{2} \ln \left (x \right )-2 y x^{2}+y^{2} \ln \left (x \right )+y^{2}}{{\mathrm e}^{x}-1}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {2 x \,{\mathrm e}^{x}-2 x -\ln \left (x \right )-1+\ln \left (x \right ) x^{4}+x^{4}-2 y \,x^{2} \ln \left (x \right )-2 x^{2} y +y^{2} \ln \left (x \right )+y^{2}}{{\mathrm e}^{x}-1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {\ln \left (x \right ) x^{4}}{{\mathrm e}^{x}-1}-\frac {2 y \,x^{2} \ln \left (x \right )}{{\mathrm e}^{x}-1}+\frac {x^{4}}{{\mathrm e}^{x}-1}+\frac {y^{2} \ln \left (x \right )}{{\mathrm e}^{x}-1}-\frac {2 x^{2} y}{{\mathrm e}^{x}-1}+\frac {2 x \,{\mathrm e}^{x}}{{\mathrm e}^{x}-1}+\frac {y^{2}}{{\mathrm e}^{x}-1}-\frac {\ln \left (x \right )}{{\mathrm e}^{x}-1}-\frac {2 x}{{\mathrm e}^{x}-1}-\frac {1}{{\mathrm e}^{x}-1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {\ln \left (x \right ) x^{4}+x^{4}+2 x \,{\mathrm e}^{x}-\ln \left (x \right )-2 x -1}{{\mathrm e}^{x}-1}\), \(f_1(x)=\frac {-2 x^{2} \ln \left (x \right )-2 x^{2}}{{\mathrm e}^{x}-1}\) and \(f_2(x)=\frac {1+\ln \left (x \right )}{{\mathrm e}^{x}-1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\left (1+\ln \left (x \right )\right ) u}{{\mathrm e}^{x}-1}} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=\frac {1}{x \left ({\mathrm e}^{x}-1\right )}-\frac {\left (1+\ln \left (x \right )\right ) {\mathrm e}^{x}}{\left ({\mathrm e}^{x}-1\right )^{2}}\\ f_1 f_2 &=\frac {\left (-2 x^{2} \ln \left (x \right )-2 x^{2}\right ) \left (1+\ln \left (x \right )\right )}{\left ({\mathrm e}^{x}-1\right )^{2}}\\ f_2^2 f_0 &=\frac {\left (1+\ln \left (x \right )\right )^{2} \left (\ln \left (x \right ) x^{4}+x^{4}+2 x \,{\mathrm e}^{x}-\ln \left (x \right )-2 x -1\right )}{\left ({\mathrm e}^{x}-1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\left (1+\ln \left (x \right )\right ) u^{\prime \prime }\left (x \right )}{{\mathrm e}^{x}-1}-\left (\frac {1}{x \left ({\mathrm e}^{x}-1\right )}-\frac {\left (1+\ln \left (x \right )\right ) {\mathrm e}^{x}}{\left ({\mathrm e}^{x}-1\right )^{2}}+\frac {\left (-2 x^{2} \ln \left (x \right )-2 x^{2}\right ) \left (1+\ln \left (x \right )\right )}{\left ({\mathrm e}^{x}-1\right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\left (1+\ln \left (x \right )\right )^{2} \left (\ln \left (x \right ) x^{4}+x^{4}+2 x \,{\mathrm e}^{x}-\ln \left (x \right )-2 x -1\right ) u \left (x \right )}{\left ({\mathrm e}^{x}-1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {d}{d x}\operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \left ({\mathrm e}^{x}-1\right )}{\left (1+\ln \left (x \right )\right ) \operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \left ({\mathrm e}^{x}-1\right )}{\left (1+\ln \left (x \right )\right ) \operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \left ({\mathrm e}^{x}-1\right )}{\left (1+\ln \left (x \right )\right ) \operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \left ({\mathrm e}^{x}-1\right )}{\left (1+\ln \left (x \right )\right ) \operatorname {DESol}\left (\left \{\frac {-2 \left ({\mathrm e}^{x}-\frac {{\mathrm e}^{2 x}}{2}-\frac {1}{2}\right ) x \left (1+\ln \left (x \right )\right ) \textit {\_Y}^{\prime \prime }\left (x \right )+\textit {\_Y}^{\prime }\left (x \right ) \left (x \ln \left (x \right )+x -1\right ) {\mathrm e}^{2 x}+\left (2 x^{3} \left ({\mathrm e}^{x}-1\right ) \ln \left (x \right )^{2}+\left (\left (4 x^{3}-x \right ) {\mathrm e}^{x}-4 x^{3}\right ) \ln \left (x \right )+\left (2 x^{3}-x +2\right ) {\mathrm e}^{x}-2 x^{3}-1\right ) \textit {\_Y}^{\prime }\left (x \right )+x \left (\left (x^{4}-1\right ) \ln \left (x \right )+x^{4}+2 x \,{\mathrm e}^{x}-2 x -1\right ) \left (1+\ln \left (x \right )\right )^{2} \textit {\_Y} \left (x \right )}{x \left ({\mathrm e}^{x}-1\right )^{2} \left (1+\ln \left (x \right )\right )}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \ln \left (x \right ) x^{4}-2 y x^{2} \ln \left (x \right )+x^{4}+y^{2} \ln \left (x \right )-2 y x^{2}+y^{2}-y^{\prime } {\mathrm e}^{x}+2 x \,{\mathrm e}^{x}-\ln \left (x \right )+y^{\prime }-2 x -1=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-2 x \,{\mathrm e}^{x}+2 x +\ln \left (x \right )+1-\ln \left (x \right ) x^{4}-x^{4}+2 y x^{2} \ln \left (x \right )+2 y x^{2}-y^{2} \ln \left (x \right )-y^{2}}{-{\mathrm e}^{x}+1} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(2*ln(x)^2*x^3+4*ln(x)*x^3+ln(x)*exp(x)*x+2*x^3+x*exp(x)-exp(x)+1)*(d Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-((ln(x)/(exp(x)-1)+1/(exp(x)-1))*y(x)^2+y(x)+(-2*x^2*ln(x)/(exp(x)-1)-2*x^2/(exp( Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] <- symmetry pattern of the form [F(x),G(x)*y+H(x)] successful <- Riccati with symmetry pattern of the form [F(x),G(x)*y+H(x)] successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 71
dsolve(diff(y(x),x) = (2*x*exp(x)-2*x-ln(x)-1+x^4*ln(x)+x^4-2*y(x)*x^2*ln(x)-2*x^2*y(x)+y(x)^2*ln(x)+y(x)^2)/(exp(x)-1),y(x), singsol=all)
\[ y \left (x \right ) = \frac {-x^{2} {\mathrm e}^{2 \left (\int \frac {\ln \left (x \right )+1}{{\mathrm e}^{x}-1}d x \right )}+c_{1} x^{2}+{\mathrm e}^{2 \left (\int \frac {\ln \left (x \right )+1}{{\mathrm e}^{x}-1}d x \right )}+c_{1}}{-{\mathrm e}^{2 \left (\int \frac {\ln \left (x \right )+1}{{\mathrm e}^{x}-1}d x \right )}+c_{1}} \]
✓ Solution by Mathematica
Time used: 2.447 (sec). Leaf size: 97
DSolve[y'[x] == (-1 - 2*x + 2*E^x*x + x^4 - Log[x] + x^4*Log[x] - 2*x^2*y[x] - 2*x^2*Log[x]*y[x] + y[x]^2 + Log[x]*y[x]^2)/(-1 + E^x),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\exp \left (\int _1^x\frac {2 (\log (K[5])+1)}{-1+e^{K[5]}}dK[5]\right )}{-\int _1^x\frac {\exp \left (\int _1^{K[6]}\frac {2 (\log (K[5])+1)}{-1+e^{K[5]}}dK[5]\right ) (\log (K[6])+1)}{-1+e^{K[6]}}dK[6]+c_1}+x^2+1 \\ y(x)\to x^2+1 \\ \end{align*}